2
$\begingroup$

I have a function $y = \sqrt{x^2 + 1} − x$, where the Domain is $(−\infty,+\infty)$.

Explanation for the domain

I need to make sure the domain of the function does not include values of $x$ that will make the square root negative.

This means that I need:

$x^2+1\ge0$

the discriminant is negative and $x \in\mathbb R$. In fact, $x^2\ge 0$, $\forall x\in\mathbb R$ and $1 > 0$.

I wish, if it's possible, to explain the value of the range with an algebraic demonstration. I am using an example to solve the function for $x$ using $y$ as parameter:

$\sqrt{x^2 + 1} − x = y$

$\sqrt{x^2 + 1} = y + x$

irrational equation, therefore:

$$ \begin{cases} & y + x >= 0\\ & x^2 + 1 =(y+x)^2 \end{cases} $$

$$ \begin{cases} & x >= -y\\ & x^2 + 1 =y^2 + x^2 + 2yx \end{cases} $$

$$ \begin{cases} & x >= -y\\ & 1 =y^2 + 2yx \end{cases} $$

$$ \begin{cases} & x >= -y\\ & x = \frac{1 - y^2}{2y} \end{cases} $$

now I need to find for which values is $x >= -y$, therefore:

$ \frac{1 - y^2}{2y} < 0$

solving the numerator and denominator

numerator

$1 - y^2 > 0$

$y^2 -1 < 0$

solve for $-1 < y < +1$

denominator

$2y > 0$

solve for $y > 0$

combine the tow solutions the inequality has occurred when

$-1 < y < 0$ or $y> 1$

$\endgroup$
  • 1
    $\begingroup$ Easier: $x^2$ is never negative, $x^2 + 1$ is even less negative. So, what is under $\sqrt\ $ is never negative. $\endgroup$ – JnxF May 9 '16 at 22:26
  • $\begingroup$ Thanks @JnxF is there a algebraic method to explain this conclusion? thanks $\endgroup$ – Gianni May 9 '16 at 22:28
  • 1
    $\begingroup$ For all $x\in\mathbb{R}$ we have that $0\leq x^2 < x^2 +1$. $\endgroup$ – Eff May 9 '16 at 22:31
  • 1
    $\begingroup$ $x^2 \geq 0 \implies x^2 + 1 \geq 1 \geq 0$. That's it. $\endgroup$ – JnxF May 9 '16 at 22:31
  • $\begingroup$ @JnxF thanks but it's still not clear sorry. I find a method to revolve the function equation for $x$ and $y$ as a parameter. I am trying to use this example but without solution. $\endgroup$ – Gianni May 9 '16 at 22:36
3
$\begingroup$

Well, there is probably an easy algebraic method: $$y=\sqrt{x^2+1}-x$$$$\implies y^2+x^2+2xy=x^2+1$$$$\implies y^2+2xy-1=0$$ to find range of $y$, we reform this equation as $$0x^2+2xy+y^2-1=0$$ this is quadratic in $x$ and since $x$ is real , $y$ should also be real so discriminent should be greater than zero $$D=(2y)^2-4(y^2-1)(0)>0$$

and this implies $$y^2>0$$$$\implies y>0$$or$$y<0$$ but $y<0$ is ruled out since $|x|=\sqrt{x^2}<\sqrt{x^2+1}$, so $x<\sqrt{x^2+1}$ ,therefore $y>0$ for all values of $x$ so $$y\in (0,\infty)$$

$\endgroup$
  • 1
    $\begingroup$ I got idea from OP $\endgroup$ – user5954246 May 9 '16 at 23:13
3
$\begingroup$

To determine the range you must study the variations of $y$ and apply the Intermediate value theorem (it's a continuous function).

Now $\;y'=\dfrac x{\sqrt{x^2+1}}-1=\dfrac {x-\sqrt{x^2+1}}{\sqrt{x^2+1}}<0,\;$ since $\;\lvert x\rvert<\sqrt{x^2+1} $. Hence the function is decreasing.

On the other hand,

  • $\lim_{x\to-\infty}\sqrt{x^2+1}-x=+\infty+\infty=+\infty$,
  • $\lim_{x\to+\infty}\sqrt{x^2+1}-x=\lim_{x\to+\infty}\dfrac{(x^2+1)-x^2}{\sqrt{x^2+1}+x}=0$.

Hence the range of $y$ is $(0,+\infty)$.

A purely computational solution:

As noted in another answer, $y=\dfrac1{x+\sqrt{x^2+1}}$. Hence the range is contained in $(0,+\infty)$. Lets show for any $m>0$, the equation $\dfrac1{x+\sqrt{x^2+1}}=m$ has a solution. Indeed, it is equivalent to $$\sqrt{x^2+1}=\frac1m-x\iff x^2+1=\Bigl(\frac1m-x\Bigr)^2\enspace\textbf{and}\enspace x \le\frac 1m.$$ Now $$x^2+1=\Bigl(\frac1m-x\Bigr)^2=\frac1{m^2}-\frac 2mx+x^2\iff x=\frac{1-m^2}{2m}$$ and the condition $\;x\le\dfrac1m$ is satisfied iff $\;1-m^2\le 2$, i.e. $m^2\ge -1$.

$\endgroup$
  • $\begingroup$ Thanks @Bernard. I am honest to say we are study to find the range of function without using the limit. (see please the example above, thanks) $\endgroup$ – Gianni May 9 '16 at 22:56
  • $\begingroup$ Then it comes down to solving quadratic equations. $\endgroup$ – Bernard May 9 '16 at 23:07
  • $\begingroup$ I posted an example but I find a strange range and probably I wrong the approach. $\endgroup$ – Gianni May 9 '16 at 23:10
  • $\begingroup$ @Gianni: I've added another solution, without using limits $\endgroup$ – Bernard May 9 '16 at 23:32
0
$\begingroup$

Hint: Expand the fraction with $\sqrt{x^2+1}+x$ to see that $$y=\frac{1}{\sqrt{x^2+1}+x}$$

$\endgroup$
0
$\begingroup$

This is only a partial answer as it only establishes the lower bound for the range.

Multiply by the conjugate

$$y=(\sqrt{x^2+1}-x) \cdot \dfrac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x}$$

$$y=\dfrac{1}{\sqrt{x^2+1}+x}$$

This function is clearly approaching $0$ as $x$ is approaching positive infinity. Also the function is never negative, since $$1>0$$

$$x^2+1>x^2$$

$$\sqrt{x^2+1}>x$$

Therefore we established a lower bound of $0$ for your range.

$\endgroup$
0
$\begingroup$

Let's see:

$$\frac1{\sqrt{x^2+1}+x}=\frac1{-x\sqrt{1+\frac1{x^2}}+x}=\frac{-\frac1x}{\sqrt{1+\frac1{x^2}}-1}$$

With l'Hospital now (we have a $\;\frac00\;$ indeterminate);

$$\lim_{x\to-\infty}\frac{-\frac1x}{\sqrt{1+\frac1{x^2}}-1}=\lim_{x\to-\infty}\frac{\frac1{x^2}}{-\frac1{x^3\sqrt{1+\frac1{x^2}}}}=\lim_{x\to-\infty}-x\sqrt{1+\frac1{x^2}}=+\infty$$

And now:

$$\frac1{\sqrt{x^2+1}+x}=\frac1{x\sqrt{1+\frac1{x^2}}+x}=\frac{\frac1x}{\sqrt{1+\frac1{x^2}}+1}\xrightarrow[x\to\infty]{}\frac02=0$$

So I'd say the range is $\;(0,\infty)\;$ as this is a continuous function everywhere.

Other way: calculate any of the two limits above, and then check that

$$\left(\sqrt{x^2+1}-x\right)'=\frac x{\sqrt{x^2+1}}-1\le0\;\;\;\forall\,x\in\Bbb R$$

so the function is monotonic decreasing. Now, if you take the limit when $\;x\to-\infty\;$ then you use this and the fact that the function's always non-negative, and if you took the limit when $\;x\to\infty\;$ then use this and the fact the function's never constant in any interval.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.